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 A124780 a(n) = gcd(A(n), A(n+2)) where A(n) = A000522(n) = Sum_{k=0..n} n!/k!. 4
 1, 2, 5, 2, 1, 2, 1, 10, 1, 2, 13, 2, 5, 2, 1, 2, 1, 10, 1, 2, 1, 2, 5, 26, 1, 2, 1, 10, 1, 2, 1, 2, 5, 2, 37, 2, 13, 10, 1, 2, 1, 2, 5, 2, 1, 2, 1, 10, 1, 26, 1, 2, 5, 2, 1, 2, 1, 10, 1, 2, 1, 2, 65, 2, 1, 2, 1, 10, 1, 2, 1, 74, 5, 2, 1, 26, 1, 10, 1, 2, 1, 2, 5, 2, 1, 2, 1, 10, 13, 2, 1, 2, 5, 2, 1, 2, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS a(n) divides n+3 because A(n+2) = (n+2)(n+1)*A(n) + n+3. LINKS Antti Karttunen, Table of n, a(n) for n = 0..4096 J. Sondow, A geometric proof that e is irrational and a new measure of its irrationality, Amer. Math. Monthly 113 (2006) 637-641. J. Sondow, A geometric proof that e is irrational and a new measure of its irrationality, arXiv:0704.1282 [math.HO], 2007-2010. J. Sondow and K. Schalm, Which partial sums of the Taylor series for e are convergents to e? (and a link to the primes 2, 5, 13, 37, 463), II, Gems in Experimental Mathematics (T. Amdeberhan, L. A. Medina, and V. H. Moll, eds.), Contemporary Mathematics, vol. 517, Amer. Math. Soc., Providence, RI, 2010. FORMULA a(n) = gcd(A000522(n), A000522(n+2)) = (n+3)/A124782(n) EXAMPLE a(2) = gcd(A(2), A(4)) = gcd(5, 65) = 5. MATHEMATICA (A[n_] := Sum[n!/k!, {k, 0, n}]; Table[GCD[A[n], A[n+2]], {n, 0, 100}]) PROG (PARI) A000522(n) = sum(k=0, n, binomial(n, k)*k!); \\ This function from Joerg Arndt, Dec 14 2014 A124780(n) = gcd(A000522(n), A000522(n+2)); \\ Antti Karttunen, Jul 07 2017 CROSSREFS Cf. A000522, A093101, A123899, A123900, A123901, A124779, A124781, A124782. Sequence in context: A270061 A061176 A180957 * A108437 A226029 A152765 Adjacent sequences:  A124777 A124778 A124779 * A124781 A124782 A124783 KEYWORD nonn AUTHOR Jonathan Sondow, Nov 07 2006 STATUS approved

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Last modified May 24 17:16 EDT 2019. Contains 323533 sequences. (Running on oeis4.)